The sum of first five multiples of 3 is:

Answer: Option A

$$S_n=\left[2a+\left(n-1\right)d\right]\times \frac{n}{2}$$
$$\Rightarrow S_5=\left[2\times 3+\left(5-1\right)3\right]$$     $$\times \frac{5}{2}$$
$$\begin{array}{rl}& \Rightarrow S_5=\left[6+12\right]\times \frac{5}{2}\\ & \Rightarrow S_5=18\times \frac{5}{2}\\ & \Rightarrow S_5=9\times 5\\ & =45\end{array}$$